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Course 2

16.532.031 Computational Electromagnetics

Prerequisites: undergraduate course on Engineering Electromagnetics or

equivalent

The required textbook is PART III of the book Electromagnetic Waves, Materials and

Computation with MATLAB, written by the instructor, to be published (August15, 2011)

by CRC Press, Taylor and Francis Group.

The text book uses the math. software MATLAB in coding the algorithms developed.

There are many examples of the MATLAB code in the text for writing the main

programs. There are five function MATLAB programs written by the instructor, whose

soft copies will be supplied by the instructor to the registrants to facilitate doing the

homework and a project. In case you are not familiar with MATLAB, the last reference is

a ‘easy to learn’, MATLAB reference. You do not have to use the MATLAB in the first 3

weeks. The homework is given so that you can do all of it it by hand computation.

Homework Problems are in the text book at the end and are listed as PX.x, X being the

chapter number and x is the serial number of the problem.. The course outline given

below lists the syllabus for the whole course. The third column gives the relevant

Sections of the hard copy for the topics listed in Column 2. The last column gives the

homework assignment.

Course Outline

16.532. Computational Electromagnetics Spring 2011

Required Text Book . Electromagnetic Waves, Materials, and Computation with

MATLAB , by Professor D. K. Kalluri published in August 2011 by CRC Press, Taylor

and Francis Group.

Wk# Date Topic Section Homework

1 01/24/2011 Formulation of E.M. Problems as Differential and Integral Equations, Finite Difference Methods, Method of Weighted Residuals

15.1 - 15.3.2

P15.2 - P15.5

2 01/31/2011 Moment Method, Finite Element Method and shape functions

15.3.3 - 15.3.5

P15.8 – P15.10

3 02/07/2011 Two Dimensional Problems: Finite Difference Method, Finite Element Method (FEM)

16.1 - 16.2

P16.1

4 02/14/2011 FEM for Poisson’s and Laplace’s Equations

16.2.1 - 16.3

P16.5

5 02/21/2011 FEM for Homogeneous Waveguide

16.4 P16.4

6 02/28/2011 First Examination

7 03/072011 FEM: second order node-based, Vector finite element techniques and edge-based shape functions

16.4.1 - 16.4.3

P16.6, P16.7

8 03/14/2011 Application of vector FEM to Homogeneous Waveguide: Characteristic impedance of Transmission Lines

16.4.4 - 16.5

P16.9,P16.10

9 03/21/2011 Moment Method, Capacitance of parallel plates taking into account fringing

16.6 P16.12

10 03/28/2011 FEM: Inhomogeneous waveguide problem; 3D-triangular elements

17.1 - 17.3.1, 17.5

Project 1

11 04/04/2011 Survey of FEM Case studies and Appendices of Chapter 16

Chapter 18, 16A - E

Project 2

12 04/11/2011 Finite Difference Time-Domain (FDTD) Method

19.1 – 19.5

P19.3, P19.4

13 04/18/2011 Continuation of FDTD and Survey of FDTD Case study

19.7 – 19.10, Chapter 20

Work on Projects

14 04/25/2011 Final Examination

15 05/02/2011 Submit Projects

Numerical Grades

There are a total of 100 points possible in the course, distributed as follows.

5 Quizes-openbook 10%

Project 1

Project 2

First Exam-openbook

Final Exam-openbook

15%

15%

30%

30%

Problems suitable for Short Quizes

P15.1, P16.2, P16.11, P16.13, P17.1

Problems suitable for Exams.

P15.6, P15.7, P16.3, P16.8, P16.14, P17.2, P19.1, P19.2

Reference Books

The text book is sufficient for the course. For completeness some references are

given below.

[1] Harrington, R. F., Field Computation by Moment Methods, Macmillan, New York,

1968.

[2] Jin Jianming, The Finite Element Method in Electromagnetics, Second Edition,

Wiley, New York, 2002.

[3] Volakis, J. L., Chatterjee, A., and Kempel, L. C., Finite Element Method for

Electromagnetics, IEEE Press, New York, 1998.

[4] Taflove, A., Computational Electrodynamics, The Finite_Difference Time_Domain,

Method, Artech House, 1995.

[5] Sadiku, N. O. M., Numerical Techniques in Electromagnetics, CRC Press, 1992.

[6] Pratap, R., Getting Started with MATLAB 5, A Quick Introduction for Scientists and

Engineers, Oxford University Press, Oxford, 1999.

Undergraduate Books

Applied Electromagnetics by Ulaby

Electromagnetic Waves by U. S. Inan & A. S. Inan

Electronic copy of the 5 function programs

% GLANT.m

function [S,T] = GLANT(Nn,Ne,n1L,n2L,n3L,xn,yn);

% Global Assembly two dimensional node based

% triangular elements

for e = 1:Ne;

n(1,e) = n1L(e);

n(2,e) = n2L(e);

n(3,e) = n3L(e);

end

% Initialization

S = zeros(Nn,Nn);

T = zeros(Nn,Nn);

% Loop through all elements

for e = 1: Ne;

% coordinates of the element nodes

for i = 1:3;

x(i) = xn(n(i,e));

y(i) = yn(n(i,e));

end

% compute the element matrix entries

b(1) = y(2) - y(3);

b(2) = y(3) - y(1);

b(3) = y(1) - y(2);

c(1) = x(3) - x(2);

c(2) = x(1) - x(3);

c(3) = x(2) - x(1);

Area = 0.5 * abs ( b(2)*c(3) - b(3) *c(2));

% Compute the elemnt matrix entries

for i = 1:3;

for j = 1:3;

Se(i,j) = (0.25/Area)*(b(i)*b(j) + c(i) *c(j));

if i ==j

Te(i,j) = Area/6;

else

Te(i,j) = Area/12;

end

% Assemble the Element matrices into Global FEM System

S(n(i,e),n(j,e)) = S(n(i,e),n(j,e))+ Se(i,j);

T(n(i,e),n(j,e)) = T(n(i,e),n(j,e))+ Te(i,j);

end

end

end

% PGLANT2.m

function [S,T,g] =

PGLANT2(Nn,Ne,n1L,n2L,n3L,Rho,Epr,xn,yn);

% Solution of Poisson's Equation

% Laplacian of V = -Rho/(epsilon0*epr)

% Rho is the array of values of volume charge density in

each element

% Epr is the array of values of dielectric constant in each

element



for e = 1:Ne;

n(1,e) = n1L(e);

n(2,e) = n2L(e);

n(3,e) = n3L(e);

end

% Initialization

S = zeros(Nn,Nn);

T = zeros(Nn,Nn);

g = zeros(Nn);


for e = 1: Ne;


for i = 1:3;

x(i) = xn(n(i,e));

y(i) = yn(n(i,e));

end


b(1) = y(2) - y(3);

b(2) = y(3) - y(1);

b(3) = y(1) - y(2);

c(1) = x(3) - x(2);

c(2) = x(1) - x(3);

c(3) = x(2) - x(1);

Area = 0.5 * abs ( b(2)*c(3) - b(3) *c(2));


ge = Rho(e)*Area/3;

for i = 1:3;

g(n(i,e)) = g(n(i,e)) + ge;

for j = 1:3;

Se(i,j) = (0.25/Area)*(b(i)*b(j) + c(i)

*c(j))*Epr(e)*8.854*10^(-12);

if i ==j

Te(i,j) = Area/6;

else

Te(i,j) = Area/12;

end

% Assemble the Element matrices into Global FEM System



end

end

end

% GLANT2T.m

function [S,T] =

GLAN2T(Nn,Ne,n1L,n2L,n3L,n4L,n5L,n6L,xn,yn);


% triangular elements of second order

% written by D. K. Kalluri

for e = 1:Ne;

n(1,e) = n1L(e);

n(2,e) = n2L(e);

n(3,e) = n3L(e);

n(4,e) = n4L(e);

n(5,e) = n5L(e);

n(6,e) = n6L(e);

end

Q1 = (1/6)*[0,0,0,0,0,0;0,8,-8,0,0,0;0,-8,8,0,0,0;

0,0,0,3,-4,1;0,0,0,-4,8,-4;0,0,0,1,-4,3];

Q2 = (1/6)*[3,0,-4,0,0,1;0,8,0,0,-8,0;-4,0,8,0,0,-4;

0,0,0,0,0,0;0,-8,0,0,8,0;1,0,-4,0,0,3];

Q3 = (1/6)*[3,-4,0,1,0,0;-4,8,0,-4,0,0;0,0,8,0,-8,0;

1,-4,0,3,0,0;0,0,-8,0,8,0;0,0,0,0,0,0];

Tek = (1/180)*[6,0,0,-1,-4,-1;0,32,16,0,16,-4;0,16,32,-

4,16,0;

-1,0,-4,6,0,-1;-4,16,16,0,32,0;-1,-4,0,-1,0,6];

% Initialization

S = zeros(Nn,Nn);

T = zeros(Nn,Nn);


for e = 1: Ne;


x(1) = xn(n(1,e));

x(2) = xn(n(4,e));

x(3) = xn(n(6,e));

y(1) = yn(n(1,e));

y(2) = yn(n(4,e));

y(3) = yn(n(6,e));


b(1) = y(2) - y(3);

b(2) = y(3) - y(1);

b(3) = y(1) - y(2);

c(1) = x(3) - x(2);

c(2) = x(1) - x(3);

c(3) = x(2) - x(1);

Area = 0.5 * abs ( b(2)*c(3) - b(3) *c(2));

COTTH1 = -(0.5/Area)*(b(2)*b(3) + c(2)*c(3));

COTTH2 = -(0.5/Area)*(b(3)*b(1) + c(3)*c(1));

COTTH3 = -(0.5/Area)*(b(1)*b(2) + c(1)*c(2));

Te = Area*Tek;

Se = COTTH1*Q1 + COTTH2*Q2 + COTTH3*Q3;


for i = 1:6;

for j = 1:6;

% Assemble the Element matrices into Global FEM

System



end

end

end

% GLAET.M

function [E,F] = glaet(Neg, Nn, Ne, n1L, n2L, n3L, n1EL,

n2EL, n3EL, xn, yn)

for e = 1:Ne

n(1,e) = n1L(e);

n(2,e) = n2L(e);

n(3,e) = n3L(e);

ne(1,e) = n1EL(e);

ne(2,e) = n2EL(e);

ne(3,e) = n3EL(e);

end

E=zeros(Neg, Neg);

F=zeros(Neg, Neg);

for e = 1:Ne

for i = 1:3;

x(i) = xn(n(i,e));

y(i) = yn(n(i,e));

end

b(1)=y(2)-y(3);

b(2)=y(3)-y(1);

b(3)=y(1)-y(2);

c(1)=x(3)-x(2);

c(2)=x(1)-x(3);

c(3)=x(2)-x(1);

Area=0.5*abs(b(2)*c(3)-b(3)*c(2));

l(1)=sqrt(b(3)*b(3)+c(3)*c(3));

l(2)=sqrt(b(1)*b(1)+c(1)*c(1));

l(3)=sqrt(b(2)*b(2)+c(2)*c(2));

for i=1:3

for j = 1:3

ff(i,j)=b(i)*b(j)+c(i)*c(j);

end

end

G(1,1)=2*(ff(2,2)-ff(1,2)+ff(1,1));

G(2,2)=2*(ff(3,3)-ff(2,3)+ff(2,2));

G(3,3)=2*(ff(1,1)-ff(3,1)+ff(3,3));

G(2,1)=G(1,2);

G(1,3)=ff(2,1)-2*ff(2,3)-ff(1,1)+ff(1,3);

G(3,1)=G(1,3);

G(2,3)=ff(3,1)-ff(3,3)-2*ff(2,1)+ff(2,3);

G(3,2)=G(2,3);

for i=1:3

for j=1:3

Ee(i,j)=(1/Area)*(l(i)*l(j));

Fe(i,j)=(1/48)*(1/Area)*l(i)*l(j)*G(i,j);

if(ne(i,e)<0)

Ee(i,j)=-Ee(i,j);

Fe(i,j)=-Fe(i,j);

else;

end;

if(ne(j,e)<0);

Ee(i,j)=-Ee(i,j);

Fe(i,j)=-Fe(i,j);

else;

end;

ane(i,e)=abs(ne(i,e));

ane(j,e)=abs(ne(j,e));

E(ane(i,e),ane(j,e))=E(ane(i,e),ane(j,e))+Ee(i,j);

F(ane(i,e),ane(j,e))=F(ane(i,e),ane(j,e))+Fe(i,j);

end

end

end

% INHWGD.m function

[Att,Btt,Btz,Bzz,C]=INHWGD(epr,mur,k0,Neg,Nn,Ne,...

n1L,n2L,n3L,n1EL,n2EL,n3EL,xn,yn);

% Inhomogeneous waveguide problem, page 73-75 text

% written by D. K. Kalluri

% Edge elements for transverse fields and node based

% elements for longitudinal fields

% Notation of text modified to remove confusion

% Text( Equation 3.39 ) My notation

% [St] Att

% [Tt] Btt

% [Sz] Bzz

% [G] Btz

% Global Assembly of two dimensional edge based


% Neg= number of edge elements

% Nn = number of nodes

% Ne = number of elements

% n1L = array containing global node number

% of the first local node of e th elemnt

% n2L = array containing global node number

% of the second local node of e th elemnt

% % n3L = array containing global node number

% of the third local node of e th elemnt

% n1EL = array containing global edge number

% of the first local edge of e th elemnt





% a negative value indicates opposite directions

% of the global edge and the local edge

for e = 1:Ne;

n(1,e) = n1L(e);

n(2,e) = n2L(e);

n(3,e) = n3L(e);

ne(1,e) = n1EL(e);

ne(2,e) = n2EL(e);

ne(3,e) = n3EL(e);

end

% Initialization

Att = zeros(Neg,Neg);

Btt = zeros(Neg,Neg);

Btz = zeros(Neg,Nn);

Bzz = zeros(Nn,Nn);


for e = 1: Ne;


for i = 1:3;

x(i) = xn(n(i,e));

y(i) = yn(n(i,e));

end


b(1) = y(2) - y(3);

b(2) = y(3) - y(1);

b(3) = y(1) - y(2);

c(1) = x(3) - x(2);

c(2) = x(1) - x(3);

c(3) = x(2) - x(1);

Area = 0.5 * abs ( b(2)*c(3) - b(3) *c(2));

l(1) = sqrt(b(3)*b(3) + c(3)*c(3));

l(2) = sqrt(b(1)*b(1) + c(1)*c(1));

l(3) = sqrt(b(2)*b(2) + c(2)*c(2));

K(1) = l(1)/(12*Area);

K(2) = l(2)/(12*Area);

K(3) = l(3)/(12*Area);


for i = 1:3;

for j = 1:3;

ff(i,j) = b(i)*b(j) + c(i)*c(j);

Se(i,j) = (0.25/Area)*(b(i)*b(j) + c(i) *c(j));

if i ==j

Te(i,j) =Area/6;

else

Te(i,j) =Area/12;

end

end;

end;

G(1,1) = 2*(ff(2,2) - ff(1,2)+ ff(1,1));

G(2,2) = 2*(ff(3,3) - ff(2,3)+ ff(2,2));

G(3,3) = 2*(ff(1,1) - ff(3,1) + ff(3,3));

G(1,2) = ff(2,3) - ff(2,2) - 2*ff(1,3) +ff(1,2);

G(2,1) = G(1,2);

G(1,3) = ff(2,1) - 2*ff(2,3)- ff(1,1) +ff(1,3);

G(3,1) = G(1,3);

G(2,3) = ff(3,1) - ff(3,3) - 2*ff(2,1)+ ff(2,3);

G(3,2) = G(2,3);

Gtz(1,1) = -K(1)*(b(1)*(b(1)-b(2))+c(1)*(c(1) -c(2)));

Gtz(2,2) = -K(2)*(b(2)*(b(2)-b(3))+c(2)*(c(2) -c(3)));

Gtz(3,3) = -K(3)*(b(3)*(b(3)-b(1))+c(3)*(c(3) -c(1)));

Gtz(1,2) =K(1)*(b(2)*(b(2)-b(1))+c(2)*(c(2) -c(1)));

Gtz(2,3) =K(2)*(b(3)*(b(3)-b(2))+c(3)*(c(3) -c(2)));

Gtz(3,1) =K(3)*(b(1)*(b(1)-b(3))+c(1)*(c(1) -c(3)));

Gtz(1,3) = -K(1)*(-y(1)*b(2)-y(2)*b(1)-y(3)*b(3)...

+x(1)*c(2)+x(2)*c(1)+x(3)*c(3));

Gtz(2,1) = -K(2)*(-y(2)*b(3)-y(3)*b(2)-y(1)*b(1)...

+x(2)*c(3)+x(3)*c(2)+x(1)*c(1));

Gtz(3,2) = -K(3)*(-y(3)*b(1)-y(1)*b(3)-y(2)*b(2)...

+x(3)*c(1)+x(1)*c(3)+x(2)*c(2));

for i = 1:3;

for j = 1:3;

Ee(i,j) = (1/Area)*(l(i)*l(j));

Fe(i,j) = (1/48)*(1/Area)*l(i)*l(j)*G(i,j);

% Assemble the Element matrices into Global FEM

System

if (ne(i,e) < 0);

Ee(i,j) = - Ee(i,j);

Fe(i,j) = -Fe(i,j);

Gtz(i,j) = -Gtz(i,j);

else;

end;

if (ne(j,e) < 0);

Ee(i,j) = - Ee(i,j);

Fe(i,j) = -Fe(i,j);

else;

end;

Atte(i,j) = (1/mur(e))*Ee(i,j) -

k0*k0*epr(e)*Fe(i,j);

Btte(i,j) = (1/mur(e))*Fe(i,j);

Btze(i,j) = (1/mur(e))*Gtz(i,j);

Bzze(i,j) = (1/mur(e))*Se(i,j) -

k0*k0*epr(e)*Te(i,j);

% for storing purposes take absolute value of

% ne(i,e) and ne(j,e) wich may have negative

values

ane(i,e) = abs(ne(i,e));

ane(j,e) = abs(ne(j,e));

Att(ane(i,e),ane(j,e)) = Att(ane(i,e),ane(j,e))+

Atte(i,j);

Btt(ane(i,e),ane(j,e)) = Btt(ane(i,e),ane(j,e))+

Btte(i,j);

Btz(ane(i,e),n(j,e)) = Btz(ane(i,e),n(j,e))+

Btze(i,j);

Bzz(n(i,e),n(j,e)) = Bzz(n(i,e),n(j,e))+

Bzze(i,j);

end

end

end

C1 = Btt - Btz*inv(Bzz)*Btz';

C2 = inv(C1);

C = C2*Att;

Courtesy of CRC Press/Taylor & Francis Group

y

b

ax

PEC, Ez = 0, that is, F = 0

ε, μ

2 2

n

n

∇t F + kc F = 0

Figure 3.1

003x001.eps


y

b

ax

PEC, = 0, that is, = 0

ε, μn

n

∂Hz∂n

∂G∂n

2 2∇t G + Kc

G = 0

~

Figure 3.2

003x002.eps


0 1 2 5

TE10

TE01

TE20

TE11

TM11

3 fc/( fc)TE10

Figure 3.3

003x003.eps


y

xε, μ

a

PEC

Figure 3.4

003x004.eps


y

xε, μ

a

PEC

α

Figure 3.5

003x005.eps


a

a

ε1

ε1

ε2

ε2

ε

ρ

Figure 3.6

003x006.eps


1.6

1.5

1.4

1.3

1.2

1.1

1.00.2 0.4 0.6 0.8 1.0 1.2 1.3

0.613

HE11

2a/λ0

β z/β

0

TE01TM01

Figure 3.7

003x007.eps

Computational ElectromagneticsWeek 2

Instructor:

1

What will be covered today

• Moment Method

• Finite Element Method and Shape Function

2

Moment Method

. . . . . . (15.65)

. . . (15.66)

3

Moment method is one of weighted residual method discussed in the previous class.Let the problem be stated as:

where L is an operator, f is the unknown function and g is excitation. The problem becomes; to find f given L and g and the boundary conditions.

Assume

gLf

N

nnn ff

1

where fn are the basic functions chosen for the problem.Choose the weight functions wm for the problem to minimize the residual. Choose the inner product for the problem.

Let us illustrate the process by giving a simple example.

Moment Method

. . . . . . (15.67a)

. . . . . . (15.67b)

4

Solve the following problem by moment method.

with the domain given as 0 < x < 1 and the boundary conditions are given by

By moment method, substitute (15.66) in (15.65), multiply by wm and form the inner product.

2

2

2

41 xdx

fd

f 0 f 1 0

The exact answer to the problem is

326

5 42 xxxxf

gwfLw m

N

nnnm ,,

1

where is the symbol for the inner product.

. . . . . . (15.67c)

. . . . . . (15.68)

Moment Method

. . . . . . (15.69)

. . . . . . (15.70)

5

For the problem, the inner product of 2 functions can be expressed as

Using (15.69) in (15.68) one obtains

(1) Pulse function P(x)

Now we need to choose the wm and fn for the problem. This needs some level of skills.Here we define two sub-sectional basic functions.

Figure 15.7 sketches this function.

1

m n m n

0

, dx

dxgwdxfLw m

N

nnnm

1

0

1

0 1

otherwise

NxxPLet

0

12

11

. . . (15.71)

xP

x

1

1

1

N

Moment Method

. . . . . . (15.72)

. . . . . . (15.73)

6

The center of the pulse function can be shifted to x = xn by defining

(2) Triangle function T(x-xn)

Figure 15.8 sketches this function.

otherwise

NxxxxP nn

0

12

11

1

10

1

111

Nxx

NxxNxxxxT

n

nnn

nxxT

1

1

2

N

xnx

Moment Method

. . . . . . (15.74)

. . . . . . (15.75)

7

After choosing wm and fn in (15.70), the result may be written as an algebraic equation

where

Equation (15.74) yields N equations for the N unknowns n.

m

N

nnmn g

1

dxLfw nmmn 1

0

1

m m

0

g w gdx . . . . . . (15.76)

We know that integration, when an impulse function is in the integral, is given by

b

j j j j

a

x x f x dx f x f a x b

0 Otherwise

. . . . . . (15.77)

Thus we can get away by choosing fn such that L fn is a sum of impulses.

Moment Method

8

This will be the result if fn =T(x-xn) and L is a second order derivative shown in Figure 15.9

Let us illustrate the calculations by choosing N =2 in (15.66). The weight functions w1 and w2, the basis functions T 1and T2 are sketched in Figure 15.10

11, f

22 , f

x

x

1

1

6

1

6

2

6

3

6

4

6

5

6

6

2

2

____ f

1

1

____ f

Moment Method

9

1 and 2 can be obtained by two algebraic equations.

The first equation is

. . . . . . (15.78)

6

4

6

223

2

12

1 xxxdx

fdLf

66

4

6

223

6/3

6/1

1

0

1111

dxxxxdxLfw

36

6

6

42

6

23

6/3

6/1

12

dxxxx

81

4041

6/3

6/1

21

0

11 dxxdxgwg

1212111 g

81

4036 21

Moment Method

10

Solving 1 and 2, we get

. . . . . . (15.80)

Similarly, we can get the second equation

81

7663 21 . . . . . . (15.79)

2222121 g

243

64,

243

5221

So the approximate solution for moment method is

6

4

243

64

6

2

243

5221 xTxTf

. . . . . . (15.81)

The accuracy can be further improved by choosing a larger value for N but the number of equations will increase as well.

Finite Element Method

11

Two aspects to the finite element method

(ii) The second aspect of the finite element method is in the technique of generating the algebraic equations.

(i) A continuous domain is broken up into a finite number of elements. Figure 15.11 shows an example. The discrete points are the vertices of the triangle.

e

The method aims to find the unknown potentials at the finite number of discrete points.

Finite Element Method – Variational Principle

12

provided the operator L is positive definite, which can be satisfied if

. . . . . . (15.82)

Instead of solving the equilibrium equation (15.65) directly, we try to find the function fthat extremizes its ‘functional’ I(f). For a functional the argument itself is a function. The functional for (15.65), in the language of linear spaces is given by

In the above X, X1, X2 are arbitrary functions that satisfy the same boundary conditions. Refer to table 15.3 for the common partial differential equations of electromagnetics and their functionals.

I f Lf , f 2 f ,g

1 2 1 2LX , X X ,LX

LX , X 0, for any X

. . . . . . (15.83)


13

Let us use an example to illustrate application of (15.82). Figure 15.12 gives a one dimensional example.

with the boundary conditions

0 V

xv

0 V

x

d

md 1

The plate dimensions are assumed to be large compared to d. Poission’s equation, which can be simplifies to the one dimensional equation

v

dx

d

2

2. . . . . . (15.85)

010

Defining the inner product for the problem as

dx1

0

2121 ,

. . . . . . (15.86)

. . . . . . (15.87)


14

The functional for (15.85) can be written as

The function (x) that minimized (15.89) is the solution of (15.85).

Integrating the first integral on the RHS of (15.88) and using the boundary conditions (15.86), we get

. . . . . . (15.89)

Notice that the functional (15.90) is the electric potential energy.

. . . . . . (15.90)

dx

xxdx

dx

dI v

1

0

1

0

2

2

2 . . . . . . (15.88)

21 1v

0 0

xdI dx 2 x dx

dx

21 1

E v

0 0

1 1 dI I dx x x dx

2 2 dx


15

The potential function (x) in each element can be expressed by using linear interpolation:

Figure 15.13 Element (e) with end points i and j. . . (15.91)

Figure 15.14 Linear interpolation

. . . (15.92)

i j e

j i

i i

j i

x x xx x

x

xix

jx

x

x

xj

xi

Another way of writing (15.91) is

jjii xNxNx

where

ij

j

ixx

xxN

ij

ij

xx

xxN

. . . . . . (15.93a)

. . . . . . (15.93b)

Ni(x) and Nj(x) are called shape functions.


16

From (15.92) and (15.93)

. . . . . . (15.98)

. . . . . . (15.99)

From (15.90)

which can be further simplified

. . . . . . (15.100)

. . . . . . (15.101)

The total functional can be obtained by summing up the functional for all the elements

ij

ij

j

ij

i

ij xxxxxxdx

d

11

11

1

dxxxNxNdx

xxI v

x

x

jjiiij

ij

x

x

ej

i

j

i

2

2

1

2

1

dxxxNdxxxN

xxI v

x

x

jj

x

x

vii

ij

ijej

i

j

i

2

2

1

elements

eE II

The algebraic equations are then obtained by minimizing (15.101) with respect to each of the unknown potentials:

k

k

EI

,0 (unknown node potential). . . . . . . (15.102)

Finite Element Method – Solve the problem

17

Assume d = 1, = 1, = 1 in Figure 15.11 and let the domain be divided into 2 elements as shown in Figure 15.16

For element (1) 1 x 2

0 1

1 2 3

2

1,0 ji xx

102 1 2 , 2

1 10 0

2 2

ji

xx

N x x N x x

i j1 20;

2 1/ 2 1./ 2

1 2

20 0

1/ 2222 2

2 2 2

0

1I 0 1 2x dx 2xdx

12 02

x0 2

2 4


18

For element (2)

1,2

1 ji xx

12

2

11

2

1

,12

2

11

1

x

xxNx

xxN ii

i j2 3; 0

21 1

22

21/ 2 1/ 2

12

22 2

2 2 21

2

01I 2 1 x dx 0 2x 1 dx

12 12

x2 x 0

2 4

1 2 22E 2

I I I 22


19

For minimum IE

E

2

2

I 10, 4 0

2

2

1

8

The exact answer for the problem is:

xx 12

1

8

1

2

1

2

1

2

1

2

1

The exact answer given in (15.105) coincides with the approximate answer by finite element method in (15.103); however it may be noted that this is not true in the entire domain. For example, (1/4) by exact answer (15.105) is

. . . . . . (15.105)

. . . . . . (15.106)

32

3

4

11

4

1

2

1

4

1

. . . . . . (15.107)


20

By finite element method we note that x=1/4 is not an end point of an elementbut it is in the domain of element(1). In this domain

The source of the error is obvious, the exact solution shows that the potential varies quadratically where as the finite element method we used assumed a linear interpolation. The shape functions are obtained based on (15.94) and are called First order shape functions. One can define second order shape functions based on quadratic interpolation.

. . . . . . (15.108)

xxNxxN ji 2,21

i i j j

1 2

x N x N x

1 2x 2x

32

2

8

1

4

120

4

21

4

1

21

1. No questions for the week 2

2. Home work: P15.6 – P15.10

Questions for the week

Electomagnetic Waves and MaterialsWeek 2

Instructor:

1

What will be covered today

• Uniform plane waves

• Good conductor approximation

• Skin effect

• Boundary condition between PEC and dielectric medium

• AC resistance of round wires

• Bounded transmission line

2

Uniform plane wavesWhat is a sourceless medium:

v = 0 and

Jsource = 0

. . . . . . (2.1)

. . . (2.2)

. . . . . . (2.3)

. . . . . . (2.4)

3

r r One-dimensional solution for a simple lossy medium with parametersin Cartesian coordinates :

0 zz HE

EH z

jkz

jkz

te

eEE

where

2 2k j

is complex.

Uniform plane wavesThe characteristic impedance is also complex

. . . . . . (2.5)

. . . (2.6)

. . . . . . (2.7)

. . . . . . (2.8)

4

Define:

where loss tangent T is defined as

2/1

j

j

k j

where, is the attenuation constant (Np/m) is the phase constant (rad/m).

Which can be obtained by solving the equation (2.4),

2/1

2 112

1

T

2/1

2 112

1

T

T

Uniform plane waves

For a low-loss dielectric, the loss tangent T << 1 and

. . . . . . (2.9)

. . . . . . (2.10)

. . . . . . (2.11)

5

1,/2

T

1, T

1,2

tan 1

T

Good conductor approximation

Definition for good conductor

. . . . . . (2.12)

. . . . . . (2.13)

6

1T

1,1

Tf

1,542

T

sZ

2)1( jjXRZ sss

is called the skin depth. The characteristic impedance or surface impedance

. . . . . . (2.14)

We can get

Skin effectWhen waves go through the good conductor, they will be attenuated exponentially as

. . . . . . (2.15)

. . . . . . (2.16)

7

For DC, f = 0, current density is uniform inside the good conductor

. . . . . . (2.17)

where

)/cos(),( /0 zteEtzE z

)/cos(),( /0 zteJtzJ z

00 EJ

for the time-harmonic AC case (f 0), we define skin depth

f/1

• The amplitude of the current density drops to e-1 (38.8%) for a distance of .

• It will drop to zero (practically) for a distance of 4.

• At high frequencies the current is practically confined to the skin of the conductor

and the phenomenon is hence described as skin effect.

B.C. between PEC and dielectric mediumFor a perfect conductor, = => = 0, we can get E = 0 Inside the conductor

. . . . . . (2.19)

8

. . . . . . (2.20)

If medium 1 is a perfect conductor and medium 2 is a dielectric, we can get the boundary conditions as

sn 212ˆ D

0ˆ212 Bn

0ˆ212 En

KH 212n

. . . . . . (2.21)

. . . . . . (2.22)

n is the normal unit vector on the boundary.

B.C. between PEC and dielectric mediumSince E = 0 Inside the conductor, we get: (2.24) and (2.26) see problem P2.2

. . . . . . (2.13)

9

. . . . . . (2.24)

. . . . . . (2.25)

. . . . . . (2.26)

0tE

0

n

En

0nH

0

n

H t

n is the normal direction and t is the tangential direction.

AC resistance

For an infinitely deep good conductor of conductivity defined by the half-space 0 < x < ,

shown in the figure, E and H can be expressed as in next slide:

10

ds

b

l

x

z

y

AC resistanceE and H inside half-space conductor

. . . . . . (2.27)

11

. . . . . . (2.28)

. . . . . . (2.29)

. . . . . . (2.30)

The time averaged power density

//0ˆ)(

~ jxxeEzxE

45//0

2ˆ)(

~ jjxx eeeEyx H

Real parts of the fields are

)(cosˆ),( /0

x

teEztx xE

)45(cos2

ˆ),( 0/0

x

teEytx xH

/220

/220

4

1ˆ45cos

22

1ˆ xx eExeEx S . . . . . . (2.31)

AC resistance

Total phasor current entering the conductor of width b

. . . . . . (2.33)

12

. . . . . . (2.35)

. . . . . . (2.32)

. . . . . . (2.36)

Define an AC equivalent resistance which consumes the same power as

After integrating, substituting the limits and transferring to time domain, we get

b

xj dydxeEddI0 0

/)1(0

~~~sEsJ

)45cos(2

)( 0

tbE

tI

Total power entering the conductor of width b and length l is given by

blEblPx

200 4

1S

PconsumedPowerRI ACRMS 2~

where 2

~ 0

bEI RMS is the AC RMS equivalent current.

AC resistance

When AC waves go through an infinitely deep good conductor of conductivity defined by

the half-space 0 < x < , shown in the figure 2.1, the equivalent DC conductor has d depth.

13

. . . . . . (2.38)

Finally, we get the AC equivalent resistance

b

lRAC

c

nonuniformdistribution

b

l uniformdistribution

b

l

=

AC resistance of round wires

The AC resistance of a round wire of

radius a and length l.

14

c

,

b

c

ac 2

,l l

R as a

1. >> a, we get

,2

lR a

a

2. << a, we get

. . . . . . (2.39)

. . . . . . (2.42)

Transmission linesA transmission line is modeled by distributed circuit theory,

15

where,

R’: The resistance due to imperfect conductors

L’: The inductance due to magnetic flux generated by the currents in the

conductors

G’: The conductance due to an imperfect dielectric

C’: The capacitance due to the surface charges on the conductors

V(z,t) V(z+z,t)

I(z,t) Rz Lz

CzGz

I(z+z,t)

z

Transmission linesThe parameters are computed as though the fields are static. For a coaxial cable shown (slide

14), the parameters are

16

. . . . . . (2.42)

. . . . . . (2.43)

. . . . . . (2.44)

. . . . . . (2.45)

For high frequencies, << a

)/ln(

2

abG

)/ln(2

abL

)/ln(

2

abC

baR

cc 2

1

2

1

Transmission linesL’ including the correction due to the internal inductance (inductance due to the magnetic flux in

the conductors) is given

17

. . . . . . (2.46)

. . . . . . (2.48)

. . . . . . (2.49)

where

For a two-port network shown in figure 2.4, the following equations can be obtained

LabL

)/ln(

2

RL

t

tzIzLtzzIRtzzVtzV

),(),(),(),(

t

tzILtzIR

z

tzV

),(),(

),(

when z -> 0

t

tzVCtzVG

z

tzI

),(),.(

),(

Similarly

. . . . . . (2.51)

Transmission linesFor a the lossless transmission line

18

. . . . . . (2.52)

. . . . . . (2.54)

. . . . . . (2.57)

where

If the source is harmonic

where

Solution

0,0v

12

2

22

2

GR

t

V

z

V

CL

1v

0~

~2

2

2

V

z

V

v

zjzj eVeVV 00

~~~

Transmission linesThe relation between the voltages and the current

19

. . . . . . (2.58)

. . . . . . (2.62)

where

For a non-PEC, we get

IZV~~

0

IZV~~

0

CLZ /0

. . . . . . (2.59)

CL

Parameters Comparison between TEM wave and transmission

TEM wave Transmission wave

E V

H I

C’

L’

Z0

CjG

LjRZ

0

Bounded Transmission linesFor a bounded transmission line,

20

where,

d: transmission line of length

Vg: Voltage source

Zg: internal impedance

ZL: External load

d

, Z0

ZL

Zg

A

B D

C

gV~

0

~~VVL Z(d)

z

IL

gI~

Bounded Transmission linesThe voltages and the current

21

. . . . . . (2.63)

. . . . . . (2.66)

where

Defined 0 as the reflection coefficient at the load

. . . . . . (2.64)

djdj eVeVdV 00

~~)(

~

djdj eVeVZ

dI 00

0

~~1)(

~

dj

dj

e

eZ

dI

dVdZ

20

20

01

1

)(~

)(~

)(

0

0

0

00 ~

~

ZZ

ZZ

V

V

L

L

By substituting (2.66) in (2.65), the input impedance can be obtained

djZdZ

djZdZZdZ

L

L

sin cos

sin cos)(

0

00

. . . . . . (2.65)

. . . . . . (2.67)

Bounded Transmission lines: see Apendice 2B and 2C for Smith Chart etc.

From 2.67, we get input impedance for special length

22

. . . . . . (2.68)

. . . . . . (2.71)

Two special cases

For a matched line, ZL = Z0 , 0 = 0

. . . . . . (2.69)

. . . . . . (2.70)

LZZZ /)4/( 20

LZZ )2/(

1,)(

d

ZdZ L

( ) .LZ d Z

)0(, tan )( 0 LZdjZdZ

)(,cot )( 0 LZdjZdZ

. . . . . . (2.72)

. . . . . . (2.73)

23

A lossfree nonuniform transmission line has

L’ = L’(z)C’ = C’(z)where L’ and C’ are per meter values of the series inductance and parallel capacitance of the transmission line.

Determine the partial differential equation for the instantaneous voltage V(z,t).For an exponential transmission lineL’(z ) = L0 exp(qz)C’(z) = C0 exp( - qz),

assuming V(z,t) = V0 exp [j(wt – kz)],determine the relation between w and k.

Questions for the week

course 2 16.532.031 computational electromagnetics … › sample › pdf ›...

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